ogeometrie: septembrie 2024

miercuri, 18 septembrie 2024

A problem with ... echo : E: 16 967 from GMB 6-7-$\infty$ /2024

The first Problem, E:15 002, appeared in GMB_4/2016, at page 212 (the password for open file is ogeometrie ). I solved it myself in the post "Rezolvarea problemelor din Gazeta Matematica (seria B) Nr 4/2016 pentru clasa a VIII-a". The official solution to the problem appeared in GMB_10/2016, at pages 473-474; I have not yet compared it with my solution...

After almost 2000 problems, another version, by the same author, appears in GMB_6-7-8/2024, at page 369 (same pasword !!).

ANSWER CiP $a=2,\;\;b=2.$

Solution CiP

We add $2+2=4$ to both sides of the equation; so

$$\left ( \frac{2a-1}{b+2}+2 \right )+\left ( \frac{2b-1}{a+2}+2 \right )=\frac{6}{a+b}+4\;\Leftrightarrow$$

$$\Leftrightarrow\;\frac{2a+2b+3}{b+2}+\frac{2b+2a+3}{a+2}=\frac{6+4a+4b}{a+b};$$

since $2a+2b+3 \neq 0$, being an odd number, we can simplify the equation with him, and get

$$\frac{1}{b+2}+\frac{1}{a+2}=\frac{2}{a+b}\Leftrightarrow \frac{1}{b+2}-\frac{1}{a+b}=\frac{1}{a+b}-\frac{1}{a+2}\Leftrightarrow $$

$$\Leftrightarrow \frac{a-2}{(b+2)(a+b)}=\frac{2-b}{(a+b)(a+2)}\;\;\;\underset{a+b \neq 0}{\Leftrightarrow} \;\;\;\frac{a-2}{b+2}=\frac{2-b}{a+2}\Leftrightarrow$$

$$\Leftrightarrow a^2-4=4-b^2\Leftrightarrow a^2+b^2=8;$$

the last equation is possible with integer numbers only if $a^2=4,\;b^2=4$ and, because $a,b \neq -2$, we have $a=2$ and $b=2.$

$\blacksquare$

marți, 17 septembrie 2024

GAZETA MATEMATICĂ Seria B N0 6-7-8/2024

This magazine is for my students, who do not have the possibility to subscribe. The password for opening the file is known only to them.

Issues of the magazine, older or newer, can be found here.

You can find other mathematics magazines here.

Sorry, the links are no longer valid.

vineri, 13 septembrie 2024

Ciąg ... podstępów // Un șir cu ... șmecherie

From the link https://www.deltami.edu.pl/2024/09/klub-44/

Statement of the problem

"The sequence $\;a_0,\;a_1,\;a_2,\;...\;$ is defined by the formulas

$a_0=3,\;a_{n+1}=a_n^2-2,\;n=0,\;1,\;2,...\;.$ Calculate the limit

$$\lim_{n \to \infty}\frac{1}{a_n}\prod_{i=0}^{n-1}a_i$$

or demonstrate that this limit does not exist."

The demonstration is from the same cited place .

ANSWER : The limit is $\frac{1}{\sqrt{5}}.$

Proof : Personal preamble CiP

A first observation is that $\;a_n>2.$ (It would be too pedantic to prove this by induction; however: $a_0>2\;$ and assuming $a_n>2\;$ we get

$a_{n+1}=a_n^2-2>2^2-2=2.$)

We have, because $a_n>2,$

$\;\;a_{n+1}-a_n=a_n^2-2-a_n=(a_n+1)(a_n-2)>0$, so $a_{n+1}>a_n$, meaning the sequence $(a_n)_{n\geqslant 0}$ is strictly increasing.

Then we have $a_n\geqslant n+3$ (an extremely poor estimate). (Induction is equally trivial: $a_0\geqslant 3$, and assuming $\;a_n \geqslant n+3$, result

$a_{n+1}\geqslant (n+3)^2-2=n^2+6n+7\geqslant n+4.$)

However, from the above it follows that

$$\lim_{n \to \infty}a_n=\infty \;. \tag{1}$$

I wrote all this to justify the statement "Jasne, że $a_n \to \infty$". Next, we copy the quoted solution.

We will define the sequence $(b_n)_{n \geqslant 0}$ by

$b_0=1,\;b_n=a_0\cdots a_{n-1},\;n=1,\;2,\dots$

and we have to investigate the limit of the sequence $\left ( \frac{b_n}{a_n}\right )_{n \geqslant 0}.$

We see that

$$b_{n+1}=a_n \cdot b_n \tag{2}$$

and we will prove by Induction that

$$a_n^2-5\cdot b_n^2=4. \tag{3}$$

Obvious $\;a_0^2-5\cdot b_0^2=3^2-5\cdot 1=4$, and assuming (3) true, result

$$a_{n+1}^2-5\cdot b_{n+1}^2 \overset{def}{\underset{(2)}{=}}(a_n^2-2)^2-5 \cdot (a_n\cdot b_n)^2=\underline{a_n^4}-4a_n^2+4-\underline{5a_n^2b_n^2}=\underline{a_n^2}\cdot (a_n^2-5b_n)-4a_n^2+4 \underset{(3)}{=}a_n^2\cdot 4-4a_n^2+4=4$$

so (3) also occurs for $\;n+1.$ This complete the Induction.

Now we have

$$\left (\frac{b_n}{a_n} \right )^2=\frac{b_n^2}{a_n^2}\underset{(3)}{=}\frac{\frac{a_n^2-4}{5}}{a_n^2}=\frac{a_n^2-4}{5a_n^2}=\frac{1}{5}-\frac{4}{5}\cdot \frac{1}{a_n^2}\xrightarrow{(1)}\frac{1}{5}-\frac{4}{5}\cdot 0=\frac{1}{5}.$$

Therefore $\;\frac{b_n}{a_n} \to \frac{1}{\sqrt{5}}.$

$\blacksquare$